Question

A ball of iron with a mass of 252 g
was heated using 2263 J
of energy. The initial temperature was 547 °C
. What was the final temperature of the iron? (The specific heat capacity for iron is 0.449 Jg°C
.) (1 point)
Responses

0.04 °C
0.04 °C

− 527 °C
− 527 °C

24.8 °C
24.8 °C

567 °C
567 °C

Answer 1

To find the final temperature of the iron ball, we can use the formula for heat transfer:

\[
Q = mc\Delta T
\]

where:
- \( Q \) is the energy supplied (in Joules)
- \( m \) is the mass of the iron (in grams)
- \( c \) is the specific heat capacity (in J/g°C)
- \( \Delta T \) is the change in temperature (final temperature - initial temperature in °C)

Given:
- \( Q = 2263 \, \text{J} \)
- \( m = 252 \, \text{g} \)
- \( c = 0.449 \, \text{J/g°C} \)
- \( T_{\text{initial}} = 547 \, \text{°C} \)

First, we rearrange the formula to solve for \( \Delta T \):

\[
\Delta T = \frac{Q}{mc}
\]

Now substituting in the known values:

\[
\Delta T = \frac{2263 \, \text{J}}{(252 \, \text{g})(0.449 \, \text{J/g°C})}
\]

Calculating the denominator:

\[
252 \, \text{g} \times 0.449 \, \text{J/g°C} = 112.548 \, \text{J/°C}
\]

Now, calculating \( \Delta T \):

\[
\Delta T = \frac{2263 \, \text{J}}{112.548 \, \text{J/°C}} \approx 20.1 \, \text{°C}
\]

Now, since \( \Delta T \) is positive, we need to add this change to the initial temperature to find the final temperature:

\[
T_{\text{final}} = T_{\text{initial}} + \Delta T = 547 \, \text{°C} + 20.1 \, \text{°C} \approx 567.1 \, \text{°C}
\]

Thus, rounded to one decimal place, the final temperature of the iron is approximately **567 °C**.

**Final Answer:** **567 °C**